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atroni [7]
3 years ago
13

A car runs around a circular race track from the point A to diametrically opposite point B The radius of the circle is ‘r’ . The

distance traveled by the car is ________ the displacement of the car is _________ show your work!
Physics
1 answer:
Anna35 [415]3 years ago
8 0

Answer:

Distance = πr

Displacement = 2r

Explanation:

First we need to find the distance covered by the car. As the car is travelling on a circular path and it traveled to a diametrically opposite point on the circular path. Therefore, the distance covered by the car must be the half value of the circumference.

Distance = Circumference/2

Distance = 2πr/2

<u>Distance = πr</u>

Since, displacement is the straight line distance between two points. So, the displacement in moving from a point to its diametrically opposite point must be equal to the diameter of circle:

Displacement = Diameter

<u>Displacement = 2r</u>

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There is a parallel plate capacitor. Both plates are 4x2 cm and are 10 cm apart. The top plate has surface charge density of 10C
liberstina [14]

Answer:

1) The total charge of the top plate is 0.008 C

b) The total charge of the bottom plate is -0.008 C

2) The electric field at the point exactly midway between the plates is 0

3) The electric field between plates is approximately 1.1294 × 10¹² N/C

4) The force on an electron in the middle of the two plates is approximately 1.807 × 10⁻⁷ N

Explanation:

The given parameters of the parallel plate capacitor are;

The dimensions of the plates = 4 × 2 cm

The distance between the plates = 10 cm

The surface charge density of the top plate, σ₁ = 10 C/m²

The surface charge density of the bottom plate, σ₂ = -10 C/m²

The surface area, A = 0.04 m × 0.02 m = 0.0008 m²

1) The total charge of the top plate, Q = σ₁ × A = 0.0008 m² × 10 C/m² = 0.008 C

b) The total charge of the bottom plate, Q = σ₂ × A = 0.0008 m² × -10 C/m² = -0.008 C

2) The electrical field at the point exactly midway between the plates is given as follows;

V_{tot} = V_{q1} + V_{q2}

V_q = \dfrac{k \cdot q}{r}

Therefore, we have;

The distance to the midpoint between the two plates = 10 cm/2 = 5 cm = 0.05 m

V_{tot} =  \dfrac{k \cdot q}{0.05} + \dfrac{k \cdot (-q)}{0.05}  = \dfrac{k \cdot q}{0.05} - \dfrac{k \cdot q}{0.05} = 0

The electric field at the point exactly midway between the plates, V_{tot} = 0

3) The electric field, 'E', between plates is given as follows;

E =\dfrac{\sigma }{\epsilon_0 } = \dfrac{10 \ C/m^2}{8.854 \times 10^{-12} \ C^2/(N\cdot m^2)} \approx 1.1294 \times 10^{12}\ N/C

E ≈ 1.1294 × 10¹² N/C

The electric field between plates, E ≈ 1.1294 × 10¹² N/C

4) The force on an electron in the middle of the two plates

The charge on an electron, e = -1.6 × 10⁻¹⁹ C

The force on an electron in the middle of the two plates, F_e = E × e

∴ F_e = 1.1294 × 10¹² N/C ×  -1.6 × 10⁻¹⁹ C ≈ 1.807 × 10⁻⁷ N

The force on an electron in the middle of the two plates, F_e ≈ 1.807 × 10⁻⁷ N

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A racing car travels with a constant tangential speed of 80.8 m/s around a circular track of radius 602 m. (a) Find the magnitud
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Answer:

10.84 m/s2 radially inward

Explanation:

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a = \frac{v^2}{r} = \frac{80.8^2}{602} = 10.84 m/s^2

b) The direction of this centripetal acceleration is radially inward

6 0
4 years ago
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